ce 208 quantity surveying

CE 208

Quantity Surveying

Department of Civil Engineering Ahsanullah University of Science and Technology

November, 2017

Department of Civil Engineering AUST

Preface

Quantity surveying refers to the estimation of materials as well as the final cost estimation for any

project. Cost estimating is one of the most important steps in project management. Cost estimation

establishes the base line of the predicted project cost at different stages of development of the

project. This lab manual intends to introduce the students with the estimation of the moderate size

structures, i.e. residential building, culvert, underground water reservoir, retaining wall as well as

estimation for excavation. The examples are imaginary structures but the basic process of the

calculations will pave the way to estimate real structures.

The authors highly indebted to their colleagues for their constant support and guidance during the

course of preparing this manual. In addition, estimation concepts were taken from Estimating

Building Costs for the Residential and Light Commercial Construction Professional, by Wayne J.

el. Pico, Estimating for Residential Construction by David Pratt. Besides, practical estimating

aspects were studied from Consultants Estimating Manual, Division of Capital Asset Management,

Commonwealth of Massachusetts (2006) while the pictures, where referenced, were collected from

the internet.

Nafis Anwari, Lecturer

Lecturer Department of Civil Engineering

Ahsanullah University of Science and Technology

Md. Hossain Nadim, Lecturer

Lecturer Department of Civil Engineering

Ahsanullah University of Science and Technology


Table of Contents Page No.

Part 1: Introduction to Quantity Surveying ......................................................................................1

1.1 Study of Drawings ...................................................................................................................1

1.2 Bill of Quantities (BoQ) ...........................................................................................................1

1.3 What is Specification? .............................................................................................................2

1.4 Necessity of Specifications .......................................................................................................2

1.5 Main Items of Works of Building: ...........................................................................................3

1.6 What is Analysis of Rate? ........................................................................................................4

1.7 Purpose of Rate Analysis .........................................................................................................5

1.8 Factors affecting the Rate analysis: .........................................................................................5

Part 2: Reinforcement Estimation of a RCC Slab .............................................................................6

2.1 Reinforcement used in the RCC ..............................................................................................6

2.2 Cover and Clear Cover in RCC ...............................................................................................6

2.3 Hook ........................................................................................................................................7

2.4 Type of Reinforcement Used in Slab ........................................................................................8

2.5 Meaning of Legends .................................................................................................................8

2.6 Length of a Cranked Bar .........................................................................................................9

2.7 Number of Reinforcement (Alternately Cranked Distribution) ............................................. 10

2.8 Worked Out Problem ............................................................................................................ 12

Part 3: Estimation of a Residential Building ................................................................................... 14

3.1 Basic Calculation of Estimation of Construction Materials ................................................... 14

3.2 Unit Weight of Some Important Materials ............................................................................ 15

3.3 Assignment ............................................................................................................................ 15


Part 4: Estimation of RCC Slab Culvert ......................................................................................... 30

4.1 Culvert .................................................................................................................................. 30

4.2 Use of Culvert ........................................................................................................................ 30

4.3 Different Types of Culvert ..................................................................................................... 31

4.4 Slab Culvert .......................................................................................................................... 33


4.5 Components of a typical RCC Slab Culvert ........................................................................... 33

4.6 Worked Out problem ............................................................................................................ 35

Part 5: Estimation of an Underground Water Reservoir ................................................................ 43


Part 6: Estimation of a Retaining Wall ........................................................................................... 47

6.1 Retaining wall ........................................................................................................................ 47

6.2 Types of Retaining wall ......................................................................................................... 47

6.3 Components of a typical RCC retaining wall ......................................................................... 48


Part 7: Estimation of a Septic Tank ................................................................................................ 53

7.1 Definition of Septic Tank ....................................................................................................... 53

7.2 Components of Septic Tank ................................................................................................... 53


Part 8: Earthwork Excavation for Roadway ................................................................................... 57

8.1 Calculation of Volume ........................................................................................................... 57

8.2 Measurement from Cross Sections ........................................................................................ 57

8.3 Terms and Abbreviations ...................................................................................................... 57

8.4 Mid-section formulae (Average Height Method) .................................................................... 58

8.5 Trapezoidal Formula/ Average End Area Method / Mean-Sectional Area Method................ 58

8.6 Prismoidal formula ................................................................................................................ 59


Part 9: Estimation of a Roof Truss .................................................................................................. 66

9.1 Types of Truss ....................................................................................................................... 66

9.2 Basic Terminologies of a truss ............................................................................................... 67


Part 10: Uses of Software in Construction Estimation .................................................................... 72

10.1 Introduction......................................................................................................................... 72

10.2 Software Used in Quantity Surveying .................................................................................. 72

10.3 QuickMeasure OS ............................................................................................................... 72

References ...................................................................................................................................... 82

CE 208: Quantity Surveying Page 1 of 83

Part 1: Introduction to Quantity Surveying

Quantity surveying is an assessment of the cost based on certain rates of materials and labour. An

estimate should be realistic assessment should be made on actual conditions of market. Estimation

is undoubtedly one or the most important aspects of a construction project. A good estimation saves

the expenditure to a great extent and ensures optimum use of materials. Estimation is very much

technical. It requires good knowledge on structural design, properly or engineering materials and

essentially practical experience.

Essentials in a good Quantity Surveyor: What are the essentials in a good Quantity Surveyor? He

must be able to describe clearly in proper unambiguous language the requirement of the Architect

and so arrange his bill of quantities (BoQ) that the Builder can quickly, easily and accurately arrive

at the estimated cost or the work. The Quantity Surveyor must have a sound knowledge or building

materials and construction and or customs prevailing in the trade. He must be accurate in his work

and calculations.

1.1 Study of Drawings

It would be extremely premature if an attempt were to be made at the taking on: immediately on

receipt of the drawings. Much has to be accomplished before proceeding to take out quantities from

the drawings. Many errors can be avoided if the following steps are taken before entertaining any

thought of taking off:

(1) Look over the drawings and attempt to visualize the work entitled.

(2) Study the, are in agreement with one another.

(3) Check drawings carefully to see that plans, elevations, sections and details, if any the dimensions

on the drawings to make sure that each overall dimension agrees with the total room dimensions. If

serious errors arc discovered, the architect should be informed; but if the discrepancy is due to the

result of slight arithmetical error, the drawings should be corrected.

(4) If the dimensions do not exist in some places, then it is always better to write them in. These

missing dimensions should be worked out from other dimensions, as far as possible, scaling them

from the drawing being the last resort.

1.2 Bill of Quantities (BoQ)

When one wants to buy a table, he ask the cabinet maker as to what is would cost. The cabinet

maker before giving the cost would like to know the detailed specification of the table, i.e., its size

length, breadth, height, design, type of timber and the finish required. Once the specification is

known the cabinet maker will give the value of the table by working out cost of materials and

labour required and adding his overhead and profit. Similarly, a prospective building owner wants

to know before placing the order that what the cost or his building would be. In order to work out

the east of a building, detailed quantities have to be worked out in accordance with the requirement



of the Standard Method of measurement and price by the Builder. These quantities when collected

together into a bill, form a bill of quantities. The advantages of a bill or quantities are as follows:

(1) It forms a common basis for competitive tendering which is necessary to obtain a reasonable

value tor consideration.

(2) It forms in itself a basis of rates for measured work which can be used in the contract for valuation

of variations and final account.

(3) It is used in building operations for the completion of interim payments.

The following points are essential in the production of good bill of quantities:

(1) A good knowledge of building construction as without this the correct interpretation of the

drawings would not be possible.

(2) Accuracy and neatness in measuring and setting out.

(3) A thorough knowledge of writing descriptions in concise and clear language which will translate

the drawings into words.

1.3 What is Specification?

A Specification is a special description or a particular subject. An engineering specification contains

detailed description or all workmanship and materials which are required to complete an engineering

project in accordance with its drawings and details. The technical drawings of a structure will

show, the proportions and relative positions of the various components of the structure. It is not

often possible to furnish the information on the drawings, regarding the quality or materials to be

used and the quality of workmanship to be achieved during construction, due to shortage of space.

This data regarding the materials and workmanship is conveyed in a separate contract document,

which is known as the "specifications" for the work. Thus the drawings with the specifications “will

completely define the structure”. The “specification” is furnished separately along with the drawings

and is an essential part of all engineering contracts.

1.4 Necessity of Specifications

The necessities of specifications are as follows:-

(a) The cost of unit quantity of work is governed by its specification.

(b) Specifications of a work are required to describe the quality and quantity or different materials

required for a construction work and is one of the essential contract documents. Thus a contractor

can make a programme to procure the materials required for a project and the owner can check

the quality of materials conforming to the specification avoiding dispute with the contractor.

(c) This also specifies the workmanship and the method of doing the work. Thus specification of a

work serves as a guide to the supervising staff of the contractor as well as to the owner to execute

the work to their satisfaction.

(d) A work is carried according to its specification and the contractor is paid for the same. Any

change in specification changes the tendered rate.



(e) As the rate or a work is based on specification, a contractor can calculate the rates of various

items of works in a tender with his procurement rates of materials and labour. Thus tender paper

without specifications of works is baseless, incomplete and invalid.

(f) Specification is necessary to specify equipment, tools and plants to be engaged for a work and

thus enables to procure them beforehand.

(g) The necessity of specification is to verify and check the strength of materials for a work involved

in a project.

(h) Specification is an essential contract document and required for arbitration of court cases

1.5 Main Items of Works of Building:

As per ASTM UNIFORMAT II Classification for Building Elements (ASTM E1557-09, 2015),

the main items of works of building are:

1. Earthwork: Earthwork in excavation and earthwork taken out accurately under different items

2. Concrete in Foundation: Foundation concrete consists of lime concrete or weak/lean cement

concrete. The proportion of cement concrete In foundation may be 1: 4: 8 or 1: 5: 10.

3. Brick Flat Soling: When the soil is soil or bad.' one layer of dry brick or stone soling is applied

below the foundation concrete. This soling layer is computed in square units specifying the

thickness.

4. Damp Proof Course: DPC usually of 2.5 cm (1 inch) thick rich cement concrete 1: 1.5:3 or 2 cm

(.75 inch) thick rich cement mortar 1 :2, mixed with standard water proofing material, is provided

at the plinth level to full width of plinth wall and the quantities are computed in square units. It

is not provided at veranda openings.

5. Masonry: Foundation and plinth masonry is taken under one item, and masonry in superstructure

is taken under a separate item. In storied building, the masonry in each storey is tabulated

separately. Proper deductions arc made for openings as doors, windows, lintels etc. Arch

masonry work is taken out separately.

6. Arch Masonry Works: Masonry work in arches is calculated in cubic units separately by

multiplying the mean length of the arch by the thickness of the: arch and by the breadth of the

wall.

7. Lintels over Openings: Length of lintel is equal to the clear span plus two i hearings. If the

dimension of bearing is not given, the bearing may be taken as 1 same as the thickness of lintel

with a minimum of 12 cm. (4.50 inch).

8. RCC & RB Work: The quantities in roof or floor slab, in beams, lintels, columns, foundations

are calculated in cubic units exclusive of steel: reinforcements and its bending but inclusive of

centering and shuttering and fixing and binding reinforcement in position. The reinforcement

including its bending is taken up separately under steel works. For this purpose, 0.6% to 1 % of

RCC by volume may be taken for steel.

9. Flooring and Roofing:

(i) Ground Floor: The base lime concrete and floor finishing of CC or stone or marble or

mosaic etc. are usually taken as one job or one item and the quantity is 'calculated in square



units. The length and breadth are the inside dimensions from wall to wall of superstructure.

(ii) lst floor, 2nd floor etc.: Supporting structure is taken separately in cubic units as RCC and

the nom finishing is taken separately in sq units as 2.5 cm or 4 cm (1 inch) or 1.5 inch) CC

or marble, mosaic etc.

10. Plastering and Pointing: Plastering usually 0.5 inch thick s calculated in square metres. While

deducting, the following rules are generally followed:

For small openings up to 0.5 sq. m (5 sq. ft) no deduction is made and at the same time no

additions are made for jambs, soffits and of sills of these openings.

For opening exceeding 0.5 sq. m (5 sq. ft) but not exceeding 3 sq m (30 sq. ft) deduction is

made for one face only and the other lace is allowed for jambs, soffits and sills which are not

taken into account separately.

For opening above 3 sq. m (30 sq. It) deduction is made for both faces of the opening, and

the jambs, soffits and sills are taken into account and added.

Plastering in ceiling usually of 12 mm (1/2 inch) thick and is computed in sq. m units under a separate

head.

11. Doors and Windows: (i) Chowkhat or Frame

(ii) Door and Window Shutters

12. Iron Work: Computed in Kg. or Quintal

13. White Washing or Colour Washing or Distempering: The quantities are computed in square units

and are usually same as for plastering. The inside is usually white-washed and the outside is

usually color-washed. The number of coals should be mentioned in the item.

14. Painting: Painting or Varnishing of doors and windows are computed in square meters, the

dimension should be taken for outer dimensions of the chowkhat i.e. the outer dimensions of

doors and windows. Painting is usually done in two or three coats, usually over a coat of priming.

Electrification and Sanitary and Water Supply Works: For Sanitary and Water Supply Works 8%

and for electrification 8% of the estimated cost of the building works is usually provided in the

estimate.

Rates: Rates or different items in the estimate are the current rates for the completion of the items

of work which include supply of materials, transport, labour, scaffolding, overheads, contractor's

profit, taxes, etc. The rates are usually taken from the PWD Schedule of Rates.

1.6 What is Analysis of Rate?

The basis of arriving at a correct and reasonable rate per unit work or supply, for a particular item

following its specification and detailed survey of materials, labour, equipment, etc. as required for

the unit work and their prevailing rates may be called as an analysis of rate.



1.7 Purpose of Rate Analysis

Main purposes of rate analysis are the following:-

(a) To determine the current rate per unit of an item at the locality.

(b) To examine the viability of rates offered by contractors.

(c) To calculate the quantity of materials and labour strength required for the project planning.

(d) To fix up labour contracts rates.

The price of an item of work is made up of the following components and a summation of these is

the rate per unit of an item.

(a) Cost of Materials

(b) Cost of Labour

(c) Cost or Equipments or Tools and Plants

(d) Cost of Overheads or Establishment Charges (including incidental)

(e) Profit

1.8 Factors affecting the Rate analysis:

The rate of an item of work mainly depends on the following factors:-

(1) Specification of the item which indicates the quality and proportion of materials, the method of

construction and protection of work.

(2) The present arte of materials for the item of work up to the worksite.

(3) Daily wages of different categories of labourer at the locality with their respective outputs. ;

(4) The range of lead and lift required for deposition of materials to carry out the item or work.

(5) Percentage charge of overheads which includes insurance and the possibility of theft or loss, etc.

(6) The, range of profit and availability of water in connection with the construction work.

Besides these the site condition, site organization and cost control during execution etc. should be

considered as these factors affect the cost per unit of work done at site.



Part 2: Reinforcement Estimation of a RCC Slab

2.1 Reinforcement used in the RCC

Two types of steel are used in RCC work. They are

1. Plain round mild steel bar

2. Deformed bar

Figure 2.1: Typical Reinforcement

Designation and cross-section area of ASTM standard reinforcing bars are given below:

Table 2.1: Diameter and Nominal Cross-Section Area of ASTM Standard Reinforcing Bars

Bar No. Diameter (in) Nominal Area (in2 )

#3 3/8 0.11

#4 4/8 0.20

#5 5/8 0.31

#6 6/8 0.44

#7 7/8 0.60

#8 8/8 0.79

#9 9/8 1.00

#10 10/8 1.27

#11 11/8 1.56

2.2 Cover and Clear Cover in RCC

Cover: Refers to distance of outer concrete surface from C.G. of steel.

Clear cover: Refers to distance of outer concrete surface from edge of steel.

Figure 2.2: Cover and Clear cover in RCC

Clear cover Co

ver

Eff

ecti

ve d

ep

th(d

)

To

tal

dep

th



Reinforcement covering is necessary for the following reasons:

1. To protect reinforcement/steel from weathering effect i.e. corrosion

2. To protect from fire.

3. The need for adequate adhesion between the steel and concrete.

4. The need to create cable and pipe channels without harming the reinforcement

According to the ACI code minimum cover should be maintained in all kind of RCC works. ACI

code provisions are given in the following table.

Table 2.2: ACI Code Provisions for Minimum Cover and Clear Cover

Beam/Column Slab

Clear cover 1.5 inch 0.75 inch

Cover 2.5 inch 1 inch

2.3 Reinforcement Hooks

In RCC work reinforcements are used in concrete to take tension. To achieve the good performance

i.e. strong bond between concrete and steel, hooks are provided at the ends of the all reinforcing bars

(especially in plain bars). Typical length of one hook is 9db to 12db, where db is the diameter of the

hook bar.

Figure 2.3: Typical 90° hook



2.4 Type of Reinforcement Used in Slab

1. Straight bar [lies at bottom]

2. Cranked bar [lies top and bottom depending on cranking]

3. Extra top [lies at top]

Figure 2.4: Typical Slab Reinforcement Pattern (straight & cranked bar).

Rules for slab reinforcement distribution

1. Start with a straight bar and end with a straight bar

2. Cranked bars are in between straight bars.

3. Extra tops are in between cranked bar.

2.5 Meaning of Legends

#3@ 6” c/c alternately cranked: #3 bars are distributed at bottom with a spacing 6”. First

bar should be straight and next bar should be cranked.

2#3 extra top in between cranked: 2#3 bars are placed in between cranked bars at top

position on one side.



Figure 2.5: Demonstration of Slab Reinforcement Pattern.

2.6 Length of a Cranked Bar

Figure 2.6: Length of a Cranked Bar

Length of straight reinforcement = L

Length of cranked reinforcement = (L– d – d) + d/cos45 +d/cos45

= (L– d – d) + 1.42d +1.42d

= L + 0.42d +0.42d

Length of cranked reinforcement = Length of straight bar + n (0.42d)

{Where, n= no. of cranked portion}

6" 6" 6" 6" 6" 6" 6" 6" 6" 6" 6" 6"

Extra Top

Straight Bar

Cranked Bar

d 45º

d

d

cos 45°

L



2.7 Number of Reinforcement (Alternately Cranked Distribution)

1. #3@ 6” c/c alt. ckd.

Figure 2.8: Straight and Cranked Reinforcement Distribution

No of straight bars = (76” - 2” - 2”) /12 +1 = 6.16 ≈ 7 Nos. (Upper rounding)

No of cranked bars = No of straight bars - 1 = 7 - 1 = 6 Nos.

6" 6"

1'

1'

76 "

30"

2"

2"



2. 2#3 extra top in between cranked

Figure 2.9: Straight, Cranked and Extra Top Distribution

No of extra top (one side) = Space available for extra top x No of extra top

= (No of cranked bar -1) x 2 = (6-1) x 2 = 10 Nos.

Table 2.3: Nominal Weight of the ASTM standard Reinforcing Bar

U.S. rebar size chart

Imperial Bar Size Metric Size (mm) Mass per unit length

lb/ft (kg/m)

#3 10 0.376 0.561

#4 13 0.668 0.996

#5 16 1.043 1.556

#6 19 1.502 2.24

#7 22 2.044 3.049

#8 25 2.670 3.982

#9 29 3.400 5.071

#10 32 4.303 6.418

#11 36 5.313 7.924

#14 43 7.650 11.41

#18 57 13.60 20.284

76 "

30"

2"

2"



2.8 Worked Out Problem

Figure 2.10: Reinforcement Detailing for a 5 inch Slab Supported on Masonry Wall



Reinforcement Estimation

Table 2.4: Calculation of Reinforcement Estimation for the Workout Problem

Bar Bar

Designation

No.

[rounded to upper 1]

Length

(ft)

Total

Length

(ft)

Lon

g D

irec

tion

#4 Straight 34′2-2 − 2"

12+ 1 = 35

50′ − 2" − 2" = 49.67′ 1738.33

#4 Cranked 35 − 1 = 34 49.67′ + 4𝑥0.42𝑥 (4

12) = 50.23′ 1707.70

#3 Extra top (34 − 1)𝑥2 = 66 1′9" − 2"+10" +22′

3= 9.75′ 643.50

#4 Extra top (34 − 1)𝑥2 = 66 22′

3+ 10" +

22′

3= 15.5′ 1023

#3 Hook 66 𝑥 2 = 132 10 𝑥3

8𝑥

1

12 = 0.3125 55.04

#4 Hook (35 + 34 + 66)𝑥2

= 270 10 𝑥

4

8𝑥

1

12= 0.417 112.59

Sh

ort

Dir

ecti

on

#3 Straight 50′ − 2" − 2"

18+ 1 = 76 34′2" − 2" − 2" = 33.83′ 2571.33

#3 Cranked 76 − 1 = 75 33.83′ + 4𝑥0.42𝑥 (3

12) = 34.13′ 2558.49

#3 Extra top 75 − 1 = 74 1′9" − 2"+10" +20′

3= 9.08′ 672.16

#4 Extra top 75 − 1 = 74 1′9" − 2"+9'+10" +

20′

3= 18.08′

1338.16

#3 Hook (76 + 75 + 74)𝑥2

= 450 10 𝑥

3

8𝑥

1

12 = 0.3125 140.62

#4 Hook 74 𝑥 2 = 148 10 𝑥4

8𝑥

1

12= 0.417 61.716

Table 2.5: Calculation of Weight of Reinforcement Bars for the Workout Problem

Bar Designation Total Length

(ft)

Weight per unit

length (lb/ft)

Weight

(lb)

#3 6642 0.376 2498

#4 5982 0.668 3996



Part 3: Estimation of a Residential Building

3.1 Basic Calculation of Estimation of Construction Materials

Find out the quantities in 100 cft cement concrete. Mix ratio (C: FA: CA= 1:2:4) and

shrinkage factor 1.5.

Volume of concrete (before mixing) = 100x1.5 = 150 cft

Required materials:

Cement= 1

1+2+4x150 = 21.428 cft =

21.428

1.25 bags = 17.142 bags ≈ 18 bags

(1 bag cement= 1.25 cft= 50 kg)

FA= 2

1+2+4x 150 = 42.856 cft

CA= 4

7x150 = 85.712 cft

Find out the quantities in 100 cft brick masonry. Factor of safety= 1.2 for wastage. Mortar

mix ratio (cement: sand) = 1:4 and shrinkage factor 1.5.

Volume of brick masonry= 100x1.2 = 120 cft

Nominal size of brick (with mortar) = 10”x5”x3”

No. of Bricks required= 120

10

12x

5

12x

3

12

= 1382.4 ≈ 1383

Volume of 1383 no. of bricks with mortar= 1383x10

12x

5

12x

3

12= 120.052 cft

Actual size of brick without mortar= 9.5”x4.5”x2.75”

Volume of 1383 no. of bricks without mortar= 1383x9.5

12x

4.5

12x

2.75

12= 94.091 cft

Volume of mortar required= 120.052 − 94.091 = 25.961 cft

Calculate quantity of cement and sand like previous problem.



3.2 Unit Weight of Some Important Materials

Table 3.1: Unit Weight of Some Important Materials

Materials Unit Wt. (lb/cft)

Brick work 120

Cement 90

RCC 150

Mild steel 490

3.3 Assignment

Question (1): Find out the ingredients required for making 20 ft x 18 ft concrete slab (thickness = 5

inch). Given, mix ratio = 1: 1.5: 3.

Answer (1): cement = 32.72727273 bags; sand (fine aggregate) = 61.36363636 cft & brick/stone

chips (coarse aggregate) = 122.7272727 cft.

Question (2): Suppose, you have 50 bags of cement. What volume of concrete you can make from

this 50 bags cement with a mix ratio of 1: 1.25: 2.50.

Answer (2): 197.9166667 cft dry concrete.

Question (3): Suppose, you have 5000 Nos. of full size brick. You want to make a brick wall.

What will be the volume of brickwork?

Answer (3): 434.0277778 cft.

Question (4): In question (3), determine the number of cement bags & volume of sand if mortar mix

ratio is 1:4.

Answer (4): 22.52604167 bags of cement; 112.6302083 cft sand.

Question (5): Given a 10 inch thick brick wall having a length of 30 feet and height of 10 feet. The

wall contains two doors (7 feet height & 3.5 feet width) and two windows (4 feet height & 6 feet

width) and a continuous lintel (25 feet length & 6 inch height). Using a factor of safety of 1.2 and

cement: sand ratio in mortar = 1: 4, Estimate the no. of bricks, no. of bags of cement & sand required

for the construction of the wall.

Answer (5): No. of bricks = 2195; cement = 9.88695 bags & sand co: 49.43475 cft.

Question (6): lime concrete (3 inch thickness) was casted over a RCC roof has a plan area or 50

feet x 30 feet. Determine its ingredients (mix ratio = 2: 2: 7).

Answer (6): Lime = 102.2727 cft; surkhi = 102.2727 cft & brick chips = 357.9545 cft.

Question (7): Find out the ingredients required for the RCC foundation up to ground level (see

Figure 4.1). Given, mix ratio = 1: 2: 4.

Answer (7): cement = 23.7193 bags (=24 bags); sand (fine aggregate) = 59.298 cft & brick/ stone

chips (coarse aggregate) = 118.59666 cft.



Figure 3.1: Plan and Elevation of Foundation for Question 7 of Assignment

Ground Level Ground Level

6'10" diameter circular column

Footing

3" C.C. Layer

3" B.F.S Layer

Elevation

14 "

14 "

10" diameter

circular column

3ʺ CC layer

3ʺ BFS layer




Estimate the materials required for the following residential building. Also find out the cost of all

materials.

Figure 3.2: Building Plan

Figure 3.3: Foundation Plan



Figure 3.4: Cross Sections of Foundations

Figure 3.5: Building Plan with Section Lines



Figure 3.6: Elevation at Section 1-1

Figure 3.7: Sunshade

Figure 3.8: Foundation Plan

3"3"

3"1'-6"

2"



Figure 3.9: Elevation at Section A-A

Figure 3.10: Elevation at Section B-B

Figure 3.11: Elevation at Section C-C



Here,

V= Volume

L= Length along center line

B= Thickness/ Width

H= Height

Earthwork excavation: (Volume)

For 25” wall foundation-

Length= 20’10”x2+22’10”x4 = 133’

Width= 2’1”+3”x2 (extra in both side) = 31”

Height= 3’3”

Volume= LxWxH= 1116.65 ft3


Length= 20’10”-31” = 18’3”

Width= 2’6”+3”x2 (extra in both side) = 36”

Height= 3’3”

Volume= L x W x H= 177.94 ft3

Column foundation-

Volume= LxWxH= (4’6”x4’6”x4’3”) x3= 258.19 ft3


Length= (9’-31”/2-4’6”/2(column footing)) x2+ (22’10”-4’6”) x2 = 47’7”

Width= 20”+3”x2 (extra in both side) = 26”

Height= 2’3”

Volume= L x W x H= 231.97 ft3

Total volume= (1116.65+177.94+258.19+231.97) = 1784.75 ft3



BFS (one layer below foundation): (Area)


Length= 20’10”x2+22’10”x4 = 133’

Width= 2’1”= 25”

Area= LxW= 277.08 ft2


Length= 20’10”-2’1” = 18’9”

Width= 2’6”= 30”


For column foundation-

Length= 4’

Width= 4’

Area= LxW= 3x (4’x4’)= 48 ft2


Length= (9’-1’3”/2 (due to 25” wall foundation)-1’/2(due to column footing)) x2+ (22’10”-1’/2-

1’/2) x2= 59’5”

Width= 1’8”= 20”


Total Area of BFS= (277.08+46.875+48+99.03) = 470.985 ft2

Cement concrete in foundation: (Volume)

To find volume of CC, multiply by the total area of BFS with thickness (3”).

Total volume of CC= 470.985 ft2x3”/12= 117.75 ft3

Brickwork in foundation (up to GL): (Volume)




20” width: 133’x20”/12x6”/12= 110.83 ft3

15” width: 133’x15”/12x6”/12= 83.12 ft3

10” width: 133’x10”/12x21”/12= 193.95 ft3

Total= 387.9 ft3


25” width: (18’9”+2.5”x2)x25”/12x6”/12= 19.97 ft3

20” width: (19’2”+2.5”x2)x20”/12x6”/12= 16.32 ft3

15” width: (19’7”+2.5”x2)’x15”/12x6”/12= 12.5 ft3

10” width: (19’7”+2.5”x2)’x10”/12x15”/12= 20.83 ft3

Total= 69.62 ft3


15” width: (59’5”+2.5”x2)x15”/12x6”/12= 37.4 ft3

10” width: (59’5”+2.5”x2)x10”/12x15”/12= 62.33 ft3

Total= 99.73 ft3

Brickwork in foundation from GL to PL: (volume)

25” wall foundation= 133’x10”x2’= 221.67 ft3

30” wall foundation= 20’x10”x2’= 33.33 ft3

20” wall foundation= ((9’-5”-5”+22’10”-5”-5”) x2) x10”x2’= 100.56 ft3

Total= 355.56 ft3

RCC in footing up to GL: (Volume)

Base slab of footing= 3x (4’x4’x1’) = 48 ft3

Column up to GL= 3x (1’x1’x2’9”) = 8.25 ft3

Total= 56.25 ft3



RCC in column from GL to PL: (Volume)

Concrete volume= 3x10”x10”x2’= 4.167 ft3

DPC (Damp Proof Course): (Volume)

Thickness= 1.5”

25” foundation= 133’x10”/12x1.5”/12 =13.6 ft3

30” foundation= 20’x10”/12x1.5”/12= 2.1 ft3

Deduct (door strip) =4’x3’4”x10”x1.5”= 1.39 ft3

BFS (one layer) in Floors: (Area)

Main room= 2 x (22’x20’) = 880 ft2

Verandah= (46’6”-20” (due to foundation below)) x (9’-10”)= 366.1 ft2

Total= 1246.1 ft2

CC (cement concrete) in Floors: (Volume)

To find volume of CC, multiply by the total area of BFS with thickness (3”).

Total volume= 1246.1 ft2 x 3”/12= 311.53 ft3

Floor finish (thickness 1/4”): (Volume)

1. Main room= 2x22’x20’x0.25”/12= 18.33 ft3

2. Door= 4x3’4”x10”/12x0.25”/12= 0.23 ft3

3. Verandah= 46’6”x9’x0.25”/12= 8.72 ft3

Deduct (due to column) = 3x10”x10”x0.25”= 0.043 ft3

4. Stair steps= 4x23’8”x1’4”x0.25”/12= 2.63 ft3

Total= 29.87 ft3



Brick wall in superstructure: (Volume)

Wall= (133’+20’) x10”x10’7”= 1349.375 ft3

Deduct:

1. Door (D1) = 4x3’4”x10”x7’= 77.78 ft3

2. Window (W1) = 8x6’x10”x4’6”= 180 ft3

3. Lintel over W1= 4x (6’+2x6”) x10”x6”= 11.67 ft3

4. Lintel over D1+W1= (22’x2+10”+2x6”) x10”x6”= 19.1 ft3

Total= 1060.825 ft3

RCC in column: (Volume)

RCC volume= 3x10”x10”x10’7”= 22.05 ft3

RCC in lintel: (Volume)

Lintel over W1= 4x7’x10”x6”= 11.67 ft3

Lintel over D1+W1= 45’10”x10”x6”= 19.1 ft3; Total= 30.77 ft3

RCC in roof: (Volume)

RCC volume= (46’6”+2x1’9”)x(30’8”+2x1’9”)x5”= 711.81ft3

LC (Lime concrete) in roof: (Volume)

LC volume= (50’-10”) x (34’2”-10”) x3” = 409.72ft3

Bricks in parapet: (Volume)

Brick= (2x (50’-5”+34’2”-5”)) x5”x2’ = 138.89 ft3

Brick work in stair: (Volume)

1. 1st step: 23’8”x10”x6” =9.86 ft3

2. 2nd step: 23’8”x20”x6” =19.72 ft3



3. 3rd step: 23’8”x30”x6” =29.58ft3

4. 4th step: 23’8”x40”x6” =39.44 ft3

Total= 98.6 ft3

R.C.C. in drop wall: (Volume)

Length= (9’-10”) x2+ (22’10”-10”) x2= 60’4”

Width= 5”

Height= 2’6”

Volume= 62.85 ft3

R.C.C. in sunshade: (Volume)

RCC Volume= 4x ((3”x6”) + (0.5x (3”+4”) x15”)) x7’= 13.71 ft3

Inside Plastering (thickness 0.25”): (Volume) Mix ratio- C:S= 1:6

1. Inside wall

a) Main room= 2x84’x (10’7”-10”skirting) x0.25”= 34.125 ft3

Deduct, Door= 4x3’4”x (7’-10”skirting) x0.25”= 2.18 ft3

Window= 8x6’x4’6”x0.25”= 4.5 ft3

Total= 27.45 ft3

b) Verandah= (22’10”x2+10”) x (10’7”-10”skirting) x0.25”= 9.45 ft3

Deduct, Door= 4x3’4”x (7’-10”skirting) x0.25”= 2.18 ft3

Window= 4x6’x4’6”x0.25”= 2.25 ft3

Droop wall= 2x5”x2’6”x0.25”= 0.043 ft3

Total= 4.98 ft3

2. Ceiling

a) Main room= 2x22’x20’x0.25”= 18.33 ft3



b) Verandah= 46’6”x9’x0.25”= 8.72 ft3

Deduct, Column= 3x10”x10”x0.25”= 0.0434 ft3

Drop wall= 60’4”x5”x0.25”= 0.524 ft3

Total= 26.48ft3

3. Edges

a) Door edges= 4x (6’2”+3’4”+6’2”) x10”x0.25”= 1.09 ft3

b) Window edges= 8x (6’x2+4’6”x2) x10”x0.25”= 2.92 ft3

Total= 4.01 ft3

4. Drop wall (inside face)

a) Inside face= 60’4”x2’6”x0.25”= 3.14 ft3

b) Bottom edge= 60’4”x5”x0.25”= 0.524 ft3

Total= 3.66 ft3

Total volume of inside plastering= 66.58 ft3

Outside plastering (thickness 0.5”): (Volume) Mix ratio- C:S= 1:4

1. GL to PL= (46’6”x2+30’8”x2) x2’x0.5”= 12.86 ft3

Deduct, Stair= 23’8”x2’x0.5”= 1.97 ft3

Total= 10.89 ft3

2. Outside wall= (46’6”+21’8”x2) x10’7”x0.5”= 39.61 ft3

Deduct, Window= 4x6’x4’6”x0.5”= 4.5 ft3

Sunshade= 4x7’x4”x0.5”= 0.39 ft3

Total= 34.72 ft3

3. Columns= 3x (10”x4) x (10’7”-10”) x0.5”= 4.06 ft3

Deduct, Drop wall= 6x (2’6”x5”x0.5”) = 0.26 ft3



Total= 3.8 ft3

4. Stairs

1st step= 2x10”x6”x0.5”= 60 in3= 0.035 ft3, 2nd step= 2x60 in3= 0.07 ft3,

3rd step= 3x60 in3= 0.1 ft3, 4th step= 4x60 in3= 0.14 ft3

Total= 0.345 ft3

5. Parapet

a) Inside= (49’2”x2+33’4”x2) x2’x0.5”= 13.75 ft3

b) Outside= (50’x2+34’2”x2) x2’x0.5”= 14.03 ft3

c) Top= (49’7”x2+33’9”x2) x5”x0.5”= 2.9 ft3

Total= 30.68 ft3

6. Sunshade

a) Bottom face= 4x (7’x1’6”x0.5”) = 1.75 ft3

b) Side edge= 4x [{(6”x3”+0.5x (3”+4”) x15”)} x2x0.5”] = 0.16 ft3

c) Front face= 4x (7’x6”x0.5”) = 0.58 ft3

d) Top face= 4x {7’x (3”+3”+15.04”) x0.5”} =2.05 ft3

e) Inside face= 4x (3”x7’x0.5”) = 0.29 ft3

Total= 4.83 ft3

7. Cornice

a) Side edge= (50’x2+34’2”x2) x5”x0.5”= 2.92 ft3

b) Bottom edge= {(30’8”+1’9”) x2+ (46’6”+1’9”) x2} x 1’9”x0.5”= 11.76 ft3

Total= 14.68 ft3

8. Drop wall

Outside face= 60’4”x2’6”x0.5”= 6.28 ft3

Total volume= 106.23 ft3



Skirting (thickness 0.75”): (Volume)

1. Main room= (20’x2+22’x2) x10”x0.75”= 4.38 ft3

Deduct, Door= 4x3’4”x10”x0.75”= 0.69 ft3

2. Verandah= 46’6”x10”x0.75”= 2.42 ft3

Deduct, Door= 4x3’4”x10”x0.75”= 0.69 ft3

3. Columns= 3x (10”x4) x10”x0.75”= 0.52 ft3

Total volume= 5.94 ft3

Costing of a residential building

Example:

Table 3.2: Calculation of Costing of a Residential Building

Item No. Item Description Quantity Price Per

Quantity Total Cost

01 Earthwork Excavation 1784.75 cft 2.13 tk/cft 3802 tk

Total Cost= X1 tk

Electrification= 8% of total cost= X2

Sanitary and water supply= 8% of total cost= X3

Estimated cost= (X1+X2+X3) tk



Part 4: Estimation of RCC Slab Culvert

4.1 Culvert

Culver is a small bridge used to carry water from one side to other. It should be covered with

embankment and composed of structural material around the entire perimeter. In general culverts

are of maximum 3 spans. Maximum span is limited to 5~6 meter. But long span culvert (>40 feet

span) was introduced in 1960’s.

4.2 Use of Culvert

1. Where natural streams intersects the roadway.

2. For passing surface drainage.

3. Bottom of depression where no natural water course.

Figure 4.1: Various Types Culverts



4.3 Different Types of Culvert

1. Arch Culvert: An arch culvert is normally a low profile culvert. It can be installed without

disturbing the causeway as it will span over the entire drainage width. They are normally

made of metal, stone masonry or RCC. They are installed easily, and you don't need to use

expensive water diversion structures to install it. Common shapes include semicircular arch,

elliptical arch, and concrete box culverts. Another benefit of these type of structure is that

the installation process will not take a lot of time, compared to traditional box culverts.

Figure 4.2: Arch Culverts (http://www.conteches.com, https://qph.ec.quoracdn.net)

2. Slab Culvert: A slab culvert is made of RCC slab. Masonry arches in culverts have

numerous problems, including difficulty in centering, shuttering, less life, more chance of

cracks and more dead load. Thus they have been replaced by simple RCC slab construction.

Slab in these culverts may be RCC or stone.

Figure 4.3: Slab Culverts (http://www.lefiltredumonde.com)

3. Pipe culvert: Pipes culverts are available in different shapes such as circular, elliptical and

pipe arches. Although circular pipes are the most common, other shapes might be used

depending on site conditions and constraints at the job site. Their prices are very competitive,

and they are very easy to install. As with other culvert types, the selection of the culvert will

depend on hydraulic design and other factors that might affect their performance and



suitability. It is the preferred one on urbanized areas and is the one usually used to manage

storm sewer systems.

Figure 4.3: Pipe Culverts (https://www.civilgeo.com)

4. Box culvert: Box culverts have a concrete (sometimes other materials can be used too) floor

allowing the water to flow smoothly through it. Box culverts are usually made up of

Reinforced Concrete (RCC). Some box culverts can be built using composite structures and

are great when water needs to change direction or when a large flow of water is expected.

Box culverts can also be installed in such way that the top of the culvert is also the roadway

surface.

Figure 4.4: Box Culverts (http://www.hudsoncivil.com.au)

5. Steel girder culvert: A steel girder culvert has two (2) steel girders running side-by-side to

support the main rail path. This type of culvert can only be seen in railways. Two main girders

are laid just below the rails. Wooden sleepers are provided between these girders and the

rails. Therefore sometimes these culverts are also called open deck culverts.



Figure 4.5: Steel Girder Culverts (http://2.bp.blogspot.com)

6. Scupper: A scupper is an opening in the side walls of an open-air structure, for purposes of

draining water. They are usually placed at or near ground level, and allow rain or liquids to

flow off the side of the open-air structure, instead of pooling within the walls.

Figure 4.6: Scuppers (http://www.nzdl.org, https://i.ytimg.com)

4.4 Slab Culvert

Slab culvert is one of the most commonly used culvert in Bangladesh. A total of 18257 culverts have

been constructed in Bangladesh until 2017 out of which 3991 culverts (22%) are box culverts

(http://www.rthd.gov.bd/bridge_maintenance.php).

Advantages of Slab Culvert

a) Simple in construction

b) Suitable for weak sub grade.

c) Uniform load distribution over a wide area.

4.5 Components of a typical RCC Slab Culvert

1. Abutment

a) Support bridge deck.

b) Retain embankment.

c) Connect approach road to bridge deck.



2. Wing wall

a) Acts as an anchor.

b) Provide smooth movement of water.

c) Prevent spilling of earthen embankment

3. Slab/deck

a) Carries the load of vehicles.

Figure 4.7: Abutment and Wing wall (oldcastleprecast-yut3re1sojoa.netdna-ssl.com)

Wing wall

Abutment



4.6 Worked Out problem

Calculate the quantity of the materials required for the construction of the following RCC slab

culvert.

Figure 4.8: Plan of the Culvert.

7800 m

m

2850

mm

2850

mm

3600 mm

Wing w

allW

ing

wal

lA

butm

ent

Abutm

ent

450 mm

450 mm

450 mm

450 mm

Wat

er F

low

Roadway



Figure 4.9: Sectional Elevation through Abutment

Figure 4.10: Sectional Elevation through Wing Wall



Concrete Estimation

Table 4.1: Calculation of Estimation of Concrete for the Workout Problem of Slab Culvert

Item Description No Length(m) Width

(m)

Height/

thickness

(m)

Area

(m2)

Volume

(m3) Remarks

(1)75mm brick

flat soling in

foundation

Abutment 2 7.80 2.325 --- 36.27 ---

Wing wall 4 2.85 2.325 --- 26.51 ---

Total= 62.78

(2) Reinforced

cement concrete

in foundation

(including

reinforcement);

1:2:4

abutment 2 7.80 2.325 0.375 --- 13.60

Wing wall 4 2.85 2.325 0.375 --- 9.94

Total= 23.54

(3) Reinforced

cement concrete

in superstructure

(including

reinforcement);

1:2:4

Abutment 2 7.80 0.45 3.825 --- 26.8515

Deduction due to

slab bearing

2 7.80 0.30 0.225 --- -1.053

Wing wall 4 2.85 0.45 3.825 --- 19.6223

Total= 45.4208

(4) RCC in slab;

1:1.5:3

1 7.80 3.60 0.225 --- 6.318

Total= 6.318



1. 75mm BFS [one layer]

Size of one brick =240 mm x115mm x 70 mm

Area of one brick =0.24 m x0.115 m = 0.0276 sqm

No of bricks = 62.78

0.0276= 2275 Nos.

Sand volume required per 10 sqm area of BFS = 0.1 cum

Volume of sand = 0.1 𝑥62.78

10 = 0.628 cum

2. Reinforced cement concrete in Foundation (1:2:4)

Final volume (hard concrete) = 23.54 cum

Initial volume (before mixing) = 23.54 x1.5 = 35.31 cum

Mix ratio = 1:2:4

Cement = 35.31 𝑥 1

7 =5.044 cum =143 bags

[One bag cement = 50 kg = 112lb. = 1.25 cft = 0.035423256 cum]

Sand = 35.31 𝑥 2

7 =10.1 cum

Brick chips/ Khoa = 35.31 𝑥 4

7 = 20.2 cum

No. of bricks = 20.2 x 300 = 6060 Nos.

[One cum brick chips required 300 Nos. of full size brick]

3. Reinforced Cement Concrete in Superstructure (1:2: 4)


Initial volume (before mixing) = 45.42 x 1.5 = 68.13 cum

Mix ratio = 1:2:4

Cement = 68.13 𝑥 1

7 = 9.7 cum = 275 bags




Sand = 68.13 𝑥 2

7 =19.5 cum


7 = 38.93 cum

No. of bricks = 38.93 x 300 = 11680 Nos.


4. RCC in Slab (1:1.5:3)


Initial volume (before mixing) = 6.32 x 1.5 = 9.48 cum

Mix ratio = 1:1.5:3

Cement = 9.48 𝑥 1

7 = 1.7 cum = 49 bags


Sand = 9.48 𝑥 1.5

7 =2.58 cum


7 = 5.17 cum

No. of bricks = 5.17 x 300 = 1551 Nos.


Total volume of cement = [143+275+49] = 467 bags

Total volume of sand = [0.628 + 10.3 + 19.5 + 2.58] =33 cum

Total number of bricks = [2275 +6060 + 11680 +1551] =21566 Nos.

The summary of quantity of materials is provided below:

Table 4.2: Summary of Quantity of Materials:

Materials Quantity

Cement 467 bags

Sand 33 cum

Brick 21566 Nos.



Reinforcement estimation:

Plain bar is used; each bar is hooked in each ends.

Hook length = nine times bar diameter = 9db

Cover = 75mm = 3 inch

Reinforcement in one wing wall:

(1) Vertical rod in wall (outside face)

19mm dia @ 125 mm C/C

= [2850−75

125] x [3600 +225 -75 +375 -75 + (12*25) + (10*19*2)]

(2) Vertical rod in wall (inside face)

12mm dia @ 150mm C/C

= [2850−75

150] x [3600 +225 -75 +375 -75 + (12*25) + (10*12*2)]

(3) Horizontal rod in wall (both face)


= [3825−75

150+ 1] x [2850 -75 + (10*10)] x 2 (two face)

(4) Horizontal rod in footing (both layer)


= [2850

150] x [1200 +450 +675 -75 -75 + (10*16*2)] x 2 (two layer)

(5) Horizontal rod in footing (both face)

10 mm dia @ 150 mm C/C

= [1200+450+675 −75−75

150+ 1] x [2850 -75 + (10*10)] x 2 (two face)



Reinforcement in one abutment:

(1) Vertical rod in wall (outside face)

19 mm dia @ 125mm C/C

= [7800

150+ 1] x [3600 + 225 +375 -75 -75 + (12*25) + (10*19*2)]

(2) Vertical rod in wall (inside face)


= [7800

150+ 1] x [3600 +375 -75 -75 + (12*25) + (10*12*2)]

(3) Horizontal rod in wall (outside face)


= [3600+225−75

150+ 1] x [7800]

(4) Horizontal rod in wall (inside face)

10mm dia @ 150 mm C/C

= [3600−75

150+ 1] x [7800]

(5) Horizontal rod in footing (both layer)

16 mm dia @150mm C/C

= [7800

150+ 1] x [1200 + 450 +675 -75 -75 + (10*16*2)] x 2 [two layer]

(6) Horizontal rod in footing (both face)

10 mm dia @150 mm C/C

= [1200+450+675 −75−75

150+ 1] x [7800] x 2 (two face)



Table 4.3: Calculation of Reinforcement in Two Abutments

Bar No

Total Length

[For one

abutment]

(m)

Weight per

meter (kg/m) Weight (kg)

19mm 2

16mm 2

12mm 2

10mm 2

Table 4.4: Calculation of Reinforcement in Four Wing Walls

Bar No

Total Length

[For one wing

wall]

(m)

Weight per

meter (kg/m) Weight (kg)

19mm 4

16mm 4

12mm 4

10mm 4

Table 4.5: Calculation of Reinforcement in two Abutments and Four Wing Walls

Bar Reinforcement

Total (kg) Two abutments (kg) Four wing walls (kg)

19mm

16mm

12mm

10mm



Part 5: Estimation of an Underground Water Reservoir


Figure 5.1: Plan of the Underground Water Reservoir



Figure 5.2: Section A-A



Concrete Estimation:

Table 5.1: Calculation of Concrete Estimation for the Workout Problem on Retaining Wall

Item Length (ft) Width (ft) Height/ Thickness

(ft) Area (ft2) Volume (ft3)

3” BFS 16 11 - 176 -

Base Slab 16 11 1 - 176

Wall 2x(9+14) 1 10-(5/12) = 9.583 - 440.83

Cover

Slab 15 10 5/12= 0.4167 - 62.5

Quantity of materials:

1. BFS

Size of one brick= 10”x5”x3”

Area of one brick= 50 in2

No. of brick required= 176x144/50= 507 Nos.

Sand per 10m2 = 0.10 m3

Volume of sand= (0.10x176x0.30482)/10= 0.1635 m3= 5.77 ft3

2. RCC (1:2:4)

Final volume (hard concrete) = 176 + 440.83 + 62.5= 679.33 ft3

Initial volume (before mixing) = 679.33 x1.5= 1018.995 ft3

Cement= (1018.995 x 1)/7= 145.57 ft3 = 117 bags

[One bag cement = 50kg = 112 lb= 1.25 ft3]

Sand= (1018.995 x 2) /7= 291.14 ft3

Brick chips= (1018.995 x 4)/7= 582.28 ft3

No. of brick= 582.28 x 0.30483 x 300= 4947 Nos.

[1m3= 35.315 ft3 brick chips required 300 Nos. of full size brick]

Table 5.2: Summary of Materials

Material Quantity

Cement 117 bags

Sand 297 ft3

Brick 5314 Nos.



Reinforcement estimation:

Table 5.3: Calculation for Reinforcement Estimation

Item Bar

designation No. Length (ft)

Total

length

(ft)

Base

slab

#6@5”c/c 2x[(16x12-2x3)/5+1]=

77

(11x12-2x3+2x10.5x6/8)/12=

11.813 909.601

#5@6”c/c 2x[(11x12-2x3)/6+1]=

44

(16x12-2x3+2x10.5x5/8)/12=

16.594 730.136

Wall

#3@5”c/c 2x[(11x12-5-2-3)/5+1]=

52 46 2392

#4@5”c/c

(outside) 46x12/5+1= 112

(11x12-5-2-

3+4+2x10.5x4/8)/12= 11.375 1274

#4@5”c/c

(inside) 46x12/5+1= 112

(11x12-5-2-

3+6+2x10.5x4/8)/12= 11.542 1292.704

Cover

slab

#6@5”c/c

(bottom) (15x12-2x3)/5+1= 36

(10x12-2x3+2x10.5x6/8)/12=

10.813 389.268

#6@5”c/c

(top) (10x12-2x3)/5+1= 24

(15x12-2x3+2x10.5x6/8)/12=

15.813 379.512

Table 5.4: Calculation for Weight of Reinforcement

Bar Length (ft) Additional

(2%)

Final Length

(ft)

Weight/length

(lb/ft)

Total weight

(lb)

#3 2392 47.84 2440 0.376 918

#4 2566.704 51.32 1619 0.668 1082

#5 730.136 14.603 745 1.043 778

#6 1678.381 33.568 1712 1.502 2572



Part 6: Estimation of a Retaining Wall

6.1 Retaining wall

A retaining wall is a structure designed and constructed to resist the lateral pressure of soil.

Generally used to protect embankment of roads, hills etc.

6.2 Types of Retaining wall

1. Gravity

2. Cantilever

3. Sheet piling

4. Anchored

5. Counterfort

Figure 6.1: Typical Retaining wall

Figure 6.2: Different Types of Retaining Wall



Figure 6.3: Different Types of Retaining Wall

6.3 Components of a typical RCC retaining wall

Figure 6.4: Components of a Typical Cantilever Retaining Wall




Calculate the quantity of the materials required for the construction of the following retaining wall.

Figure 6.5: Section of a Retaining Wall



Length of wall =300 ft

Cover for all sides = 3 in

Estimation of concrete (1:2:4)

Volume of wall = 1

2 x (1.5′+3′) x 20′ x 300′ =13500 cft

Volume of base = 8′ x 2′ x 300′ = 4800 cft

Volume of key = 1

2 x (0.5′+1′) x 1′ x 300′ = 225 cft

Total volume of concrete =18525 cft

Total volume of wet concrete =1.5 x 18525 =27787.5 cft

Table 6.1: Summary of Materials

Materials Volume (cft) Quantity

Cement 1 x 27787.5

7 = 3969.6 3176 bags

Fine Aggregate(sand) 2 x 27787.5

7 = 7939.3 7939.3 cft

Coarse Aggregate

( brick chips)

4 x 27787.5

7 =15875.6 450 Nos.



Estimation of Reinforcement

a) Reinforcement in wall

Inside vertical reinforcement (# 6 @ 5″c/c)

= (300′x12-3″-3″

5″ +1) x (20′x12+2′x12-3″-3″+2 x 9.5 x

6

8 ″) /12 = 16330.5 fOutside vertical

reinforcement (# 4 @ 7″c/c)

From figure

20.05′

20′ =

𝐿

21.75=> L=21.8′

Figure 6.6: Calculation of Inclined Length of Outside Vertical Reinforcement

= ( 300′x12-3″-3″

7″ +1) x (21.8′x12 -3″+2 x 9.5 x

4

8 ″) /12 = 11493.2 ft

Inside horizontal reinforcement (# 5 @ 6″c/c)

= ( 22′x12-3″-3″

6″ +1) x (300′x12 -3″-3″+2 x 9.5 x

5

8 ″) /12 = 13221.5 ft

Outside horizontal reinforcement (# 4 @ 8″c/c)

= ( 21.8′x12-3″

8″ +1) x (300′x12 -3″-3″+2 x 9.5 x

4

8 ″) /12 = 10007.2 ft

b) Reinforcement in base

Along length of wall (Top) (# 6 @ 6″c/c)

= ( 8′x12-3″-3″

6″ +1) x (300′x12 -3″-3″+2 x 9.5 x

6

8 ″) /12 = 4811 ft

1.5' 1.5'

2'- 312'

20'

2'

1.5'

20.05'L



Along length of wall (Bottom) (# 5 @ 7″c/c)

= ( 8′x12+-3″-3″

7″ +1) x (300′x12 -3″-3″+2 x 9.5 x

5

8 ″) /12 = 4163.9 ft

Along width of wall (Top) (# 5 @ 6″c/c)

= ( 300′x12 -3″-3″

6″ +1) x (8′x12 -3″-3″+2 x 9.5 x

5

8 ″) /12 = 5093.8 ft

Along width of wall (Bottom) (# 6 @ 7″c/c)

= ( 300′x12 -3″-3″

7″ +1) x (8′x12 -3″-3″+2 x 9.5 x

6

8 ″) /12 = 4469.1 ft

c) Reinforcement in key

From figure

3

12 =

X

15 => X= 3.75″

Along length of wall (2#8 bar)

=2 x (300′x12 -3″-3″+2 x 9.5 x 8

8 ″) /12 =602.2 ft

Figure 6.7: Reinforcement in Key

Dowel bar (# 6 @ 7″c/c)

=2 x (300′x12 -3″-3″

7″ +1) x (√ (152+3.752) ″-3″+2 x 9.5 x

6

8 ″) /12 = 2290.2 ft

Table 6.2: Calculation of Weight of Reinforcement

Bar Total length (ft) Weight/ length (lb/ft) Weight (lb)

#4 bar 11493.2 +10007.2 = 21501 0.668 14363

#5 bar 13221.5+4163.9+5093.8=22480 1.043 23447

#6 bar 16330.5+4811+4469.1+2290.2=27901 1.502 41908

#8 bar 603 2.670 1611

X

12"

3"

6"

12"

3"



Part 7: Estimation of a Septic Tank

7.1 Definition of Septic Tank

A septic tank is a watertight chamber made of concrete, fiberglass, PVC or plastic, through which

domestic wastewater (sewage) flows for primary treatment.[1] Settling and anaerobic processes

reduce solids and organics, but the treatment is only moderate.[1] Septic tank systems are a type

of onsite sewage facility (OSSF). They can be used in areas that are not connected to

a sewerage system, such as rural areas. The treated liquid effluent is commonly disposed in a septic

drain field which provides further treatment.

7.2 Components of Septic Tank

1. Inspection Pit: A hole in the ground, lined with site built or manufactured sides that receive

wastewater from house or building. The wastewater then flows from inspection pit to septic

tank.

2. Septic Tank: The septic tank is buried, watertight container typically made of concrete,

fiberglass, or polyethylene. It holds the wastewater long enough to allow solids to settle out,

forming sludge, and oil and grease to float to the surface as scum. It also allows partial

decompositions of the solid materials. Compartments and a T-shaped outlet in the septic tank

prevent the sludge and scum from leaving the tank and traveling into the soak pit.

3. Soak Pit: A soak pit, also known as a soak away or leach pit, is a covered, porous-walled

chamber that allows water to slowly soak into the ground. Pre-settled effluent from a

collection and storage/treatment or (semi-) centralized treatment technology is discharged

to the underground chamber from which it infiltrates into the surrounding soil.

Figure 7.1: Components of a Septic Tank (http://www.natureclean.com)



The following figure is related to workout problem. The septic tank has a total height of 9 ft (including 6ʺ floor slab and 6ʺ cover

slab). Soak pit has a total height of 30 ft (including 4ʺ cover slab).

INSPECTION PIT SEPTIC TANK SOAK PIT

Figure 7.2: Plan of Septic Tank

6"

10"

1'

8"

1'

8"

1'

8"

1'

6'

5"

3"10" 2' 10"

3"2' 6" 10" 5" 7'-7" 5" 4'-7" 5" 10" 6" 2'

3"3'-6"

3"


Table 7.1: Estimation of a Septic Tank

Item

No.

No. Item Description No. Length (ft) Width(ft)

Height/

Depth(ft) Quantity Remarks

1

Earthwork Excavation

Septic Tank 1 16.0833 8.67 9.0 1254.54 cft

Soak Pit 1 12.57 --- 30.0 376.99 cft

2 Cement Concrete (1 :3:6) Floor

of Septic Tank 1 16.0833 8.67 0.5 69.72 cft

3

Precast RC Work

Roof Cover Slab of Septic Tank 1 15.0833 7.67 0.5 57.84 cft

Roof Cover Slab of Soak Pit 1 12.57 --- 0.33 4.19 cft

Side Wall of Soak Pit --- 2.94 --- 29.67 87.22 eft

1st Class Brickwork with 1 :4

Cement Mortar in septic tank

(a) Long Wall

: 1st step 2 15.0833. 1.25 3.0 113.12 cft

: 2nd step 2 15.0833 0.833 5.0 125.69 cft

(b) Short Wall

: 1st step 2 5.167 1.25 3.0 38.75cft

: 2nd step 2 6.0 0.833 5.0 49.98 cft

(c) Partition Wall

: 1st step 1 5.167 0.417 3.0 6.46 cft

: 2nd step 1 6.0 0.417 5.0 12.51 cft

5 1/2 inch Cement Plaster 1:3 with

Standard Water Proofing

Compound in Septic Tank

(a) Long Wall

: 1st step 2 12.583 --- 3.0 75.498 sft

: 2nd step 2 13.417 --- 5.0 134.17 sft

(b) Short Wall

: 1st step 2 5.167 --- 3.0 31.002 sft

: 2nd step 2 6.0 --- 5.0 60.00 sft

(c) Partition Wall

: 1st step --- 5.167 --- 3.0 31.002 sft

: 2nd step --- 6.0 --- 5.0 60.00 sft

6 3/4 inch Cement Plaster 1:3

with standard water proofing

compound in floor of septic tank

1 12.583 5.167 --- 65.02 sft



Table 7.2: Cost Estimation of Septic Tank

Item

No.

Item Description Quantity times price

per quantity

Total Price (Taka)

1 Earthwork Excavation 1631.53 cft @ 1475.00

per cft

2406.50

2 Cement Concrete (1 :3:6)

69.72 cft @ 8991.00 per

cft

6268.52

3 Precast RC Work 149.25 cft @ 15073.53

per cft

22497.24

4 1st Class Brickwork with 1:4 Cement Mortar in

septic tank

346.51 cft @ 5629.40

per cft 19506.43

5 1/2 inch Cement Plaster 1 :3 with Standard Water

Proofing Compound in Septic Tank 391.67 sft @ 882.00 per

sft 3454.53

6 3/4 inch Cement Plaster 1:3 with Standard Water

Proofing Compound Floor of in Septic Tank

65.02 sft @ 1323.00 per

sft 860.21

7 Aggregate at Bottom of Soak Pit 9.62 cft @ 33.50 per cft 322.27

8 Coarse Sand at Bottom of Soak Pit 14.43 cft @ 33.50 per cft 483.40

9 2 inch dis Ventilating Pipe fitted position 1 No. @ Tk 15.00 each 15

10 6 inch diameter Pipe 5.25 ft @ Tk 30 per ft 157.50

11 C. I. (cast iron) Manhole Cover 18 inch diameter over

Septic Tank 2 No. @Tk. 300.00 each 600

12 RCC Tees 2 No. @Tk. 150.00 each 300

Item

No.

No. Item Description No. Length (ft) Width(ft)

Height/

Depth(ft) Quantity Remarks

7 2 inch size Brick Aggregate at

Bottom of Soak Pit --- π×(3.5)2/4 --- 1.0 9.62 sft

8 Coarse Sand at Bottom of Soak

Pit --- π×(3.5)2/4 --- 1.5 14.43 sft



Part 8: Earthwork Excavation for Roadway

8.1 Calculation of Volume

There are three methods generally adopted for computation of earthwork volume (according to the

formation of the solid). They are:

1) From cross sections: Measurement from cross section is a universally applicable method.

2) From spot levels: measurement from spot levels are applied sometime for large excavation.

3) From contours: Rough estimates of volume may be made by treatment of the contour line

and not much used in practice.

8.2 Measurement from Cross Sections

The cross sectional area along the line is first calculated by standard formulae and the volumes of

the prismoids between successive cross-sections are then calculated by following methods:

1) Formulae of Mid-section method/ Average height method.

2) Formulae of Trapezoidal method/ Average end area method/ mean-sectional area

method.

3) Formulae of prismoidal method according to Simpson’s one-third rule.

8.3 Terms and Abbreviations

EGL (Existing Ground Level) or GL (Ground Level): The existing earth surface

FL (Formation Level): The proposed level of roadway.

RL (Road Level): A level stated in relation to a known bench mark or datum.

Longitudinal Slope/ Gradient: Gradient may be defined as the rate of rise or fall along the length of

highway.

Side Slope: Side slope is defined as the rate of rise or fall of the shoulders of the pavement. It depends

on the soil characteristics and geographic location of the highway.



8.4 Mid-section formulae (Average Height Method)

Figure 8.1: Cross Section of a Trapezoidal Section for Average Height Method

Depth section (1) = d1 (note that d1 is the difference between GL & FL)


Average depth, davg = (d1 + d2)/2

Width of section = b

Side slope = 1: s (vertical: horizontal)

Area of mid-section, A mid = bdavg + (1/2) sdavg2 + (1/2) sdavg2

A mid = (b+sdavg) x davg

Length between two consecutive sections (between section (1) & (2)) = L1-2

Volume of earthwork between these two consecutive sections (between section (1) & (2)), V1-2 =

Aavg x L

V1-2 = (b+sdavg) x davg x L (may be cut or fill)

8.5 Trapezoidal Formula/ Average End Area Method / Mean-Sectional Area Method



Area at end 1, A1 = (b+sd1) x d1


Mean sectional area, A mean = (A1+A2)/2



d1

d2

b

d

avg

L



Length between two consecutive sections (between section (1) & (2)) = L

Volume of earthwork between these two consecutive sections (between section (1) & (2)), V1-2 = A

mean x L

V1-2 = A mean x L (may be cut or fill)

8.6 Prismoidal formula





Mean sectional area, A mean = (A1+A2)/2



Length between two consecutive sections (between section (1) & (2)) = L

Volume of earthwork between these two consecutive sections (between section (1) & (2)),

V1-2 = (A1+ 4Am + A2)/6 x L (may be cut or fill)


A 1 km road is to be constructed in existing ground level having reduced levels 54.1, 53.8, 53.5,

53.5, 54.3, 54.6, 54.9, 54.5, 54.7 and 54.3 meters at 100 m intervals. A required reduced level at

station 1 is 55 meter and the downward gradient is 1 in 1000. The width of the road at formation

level is 8 meter. Slopes to be maintained at cutting and filling are 1:2 (V: H) and 1:3 (V: H)

respectively. Calculate the volume of Earthwork.

Figure 8.2: Typical Fill Section



5 6

RL = 54.6 m

RL = 54.5 m

RL = 54.6 m

RL = 54.3 m

X 100 - X

100 m

0.3 m

0.1 m

Figure 8.3: Typical Cut Section

Figure 8.4: Long Section of the Road

*From figure

X

0.3 =

100 - X

0.1 => X= 75’

Figure 8.5: Length of Cut and Fill between station 5 & 6.



Table 8.1: Earthwork Computation Table (Mid-section / average height method)

Station FL

(m)

EGL

(m)

Depth,

d= EGL

῀ FL

(m)

Average

depth,

davg

(m)

Area,

A=(b+sdavg)davg (m2)

Length,

L (m)

Volume,

V = A x L

(m3)

Remark

1 55.0 54.1 0.9

1 (8+3 X 1) X 1

= 11 100 1100.00 Fill

2 54.9 53.8 1.1

1.2 (8+3 X 1.2) X 1.2

= 13.92 100 1392.00 Fill

3 54.8 53.5 1.3

1.25 (8+3 X 1.25) X 1.25

= 14.68 100 1468.00 Fill

4 54.7 53.5 1.2

0.75 (8+3 X 0.75) X 0.75

= 7.68 100 768.00 Fill

5 54.6 54.3 0.3

0.15 (8+3 X 0.15) X 0.15

= 1.26 75* 95.06 Fill

0 x x 0

0.05 (8+2 X 0.05) X 0.05

= 0.40 25* 10.18 Cut

6 54.5 54.6 0.1

0.15 (8+2 X 0.15) X 0.15

= 1.245 100 124.50 Cut

7 54.4 54.6 0.2

0.4 (8+2 X 0.4) X 0.4

= 3.52 100 352.00 Cut

8 54.3 54.9 0.6

0.45 (8+2 X 0.45) X 0.45

= 4.01 100 400.50 Cut

9 54.2 54.5 0.3

0.45 (8+2 X 0.45) X 0.45

= 4.01 100 400.50 Cut

10 54.1 54.7 0.6

0.45 (8+2 X 0.45) X 0.45

= 4.01 100 400.50 Cut

11 54.0 54.3 0.3

Volume of total cutting = 1688.18 m3

Volume of total filling = 4824.56 m3



Assignment 1: Calculate the volume of cutting and filling for the previous worked out problem

using the trapezoidal method.

Hints

Table 8.2: Earthwork Computation Table (Trapezoidal formula/Average End Area Method)

Sta

tion

FL

(m)

EG

L(m

)

Dep

th, d

=

EG

L F

L

(m)

Are

a , A

= (

b+

sd)d

Avg. A

rea

Am

ea

n

(m2

)

Len

gth

, L

(m

)

Volu

me

V=

Am

idxL

(m3)

Rem

ark

s

1 55 54.1 0.9

2 54.9 53.8 1.1

3 54.8 53.5 1.3

4 54.7 53.5 1.2

5 54.6 54.3 0.3

0 - - 0

6 54.5 54.6 0.1

7 54.4 54.6 0.2

8 54.3 54.9 0.6

9 54.2 54.5 0.3

10 54.1 54.7 0.6

11 54 54.3 0.3



Assignment 2: Calculate the volume of cutting and filling for the previous worked out problem

using the prismoidal method.

Hints

Table 8.3: Earthwork Computation Table (Prismoidal formula)

Sta

tion

FL

(m)

EG

L(m

)

Dep

th, d

= E

GL

FL

(m)

Are

a (

A)

Avg. d

epth

, d

avg

(m)

Are

a

Am

id=

(b+

sdav

g)

dav

g

(m2

)

Len

gth

, L

(m

)

Volume

V=(A1 +4

Amid + A2

) x L /6

(m3)

Rem

ark

s

1 55 54.1 0.9 A1

A1-2

2 54.9 53.8 1.1 A2

A2-3

3 54.8 53.5 1.3 A3

A3-4

4 54.7 53.5 1.2 A4

A4-5

5 54.6 54.3 0.3 A5

A5-0

0 - - 0 0

A0-6

6 54.5 54.6 0.1 A6

A6-7

7 54.4 54.6 0.2 A7

A7-8

8 54.3 54.9 0.6 A8

A8-9

9 54.2 54.5 0.3 A9

A9-10

10 54.1 54.7 0.6 A10

A10-11

11 54 54.3 0.3 A11



Assignment 3: Calculate the volume of earth filling for the ditch. Given, reduced level of filling

plane is 100 m.

Figure 8.6: Section A-A of Ditch for Assignment 3



Reduced levels (RL) of different stations are given as follows

Table 8.4: Reduced levels (RL) of Different Stations for Assignment 3

Station RL (m) Station RL (m) Station RL (m) Station RL (m) Station RL (m)

1 101.00 12 103.20 23 105.25 34 104.14 45 101.00

2 102.00 13 103.90 24 107.00 35 100.00 46 103.20

3 101.70 14 101.30 25 108.36 36 101.10 47 101.10

4 102.00 15 100.16 26 105.90 37 104.80 48 102.15

5 101.00 16 104.15 27 103.00 38 105.60 49 102.00

6 100.60 17 106.85 28 100.00 39 103.00

7 100.10 18 107.65 29 101.70 40 107.80

8 100.60 19 105.90 30 104.80 41 104.14

9 106.00 20 104.00 31 105.60 42 101.00

10 107.50 21 102.00 32 103.00 43 100.10

11 105.10 22 101.70 33 107.8 44 102.30

Hints:

Calculate depth to be filled at each station

Calculate volume of each grid to be filled ( i.e. V 1-2-8-9 = 1/4 x (d1 + d2 + d8 + d9) x A 1-2-8-9

)

Sum up all to get total volume (Total volume = V 1-2-8-9 + V 2-3-9-10 + …………….)



Part 9: Estimation of a Roof Truss

A truss is a structure composed of slender members joined together at their end points. Planar trusses

lie in a single plane. Typically, the joint connections are formed by bolting or welding the end

members together to a common plate, called a gusset plate. The basic building block of a truss is a

triangle. Large truss are constructed by attaching several triangles together. A new triangle can be

added truss by adding two members and a joint. A truss constructed in this fashion is known as a

simple truss.

9.1 Types of Truss

Various types of trusses are used around the world. Some of them are shown below:

Towne Lattice Truss

Pratt Truss

Truss with monitor

Bowstring Truss



Pratt Truss

Pratt Truss

Howe Truss

Warren Truss

Figure 9.1: Different Types of Truss

9.2 Basic Terminologies of a truss

Bay: Bay is the distance between two main trusses.

Span: Span is the distance between supports of a truss.

Rise: The rise of a truss is vertical distance between apex and line joining the supports.

Pitch: The ratio of the rise to the span is called the pitch.

Top Chord: The uppermost line of members which extend from one support to the other through the apex is

called the upper chord.

Bottom Chord: The bottom chord consists of the lowermost line of members extending from one support to

other.

Vertical: Vertical members connecting top and bottom chords.

Diagonal: Inclined members connecting top and bottom chords.



Figure 9.2: Dimensions of a Truss System

Figure 9.3: Components of a Truss

Rise

Span

Bay

Bay

Sagrod

Purlin

Top Chord

Vertical

Bottom Chord Diagonal



Figure 2: Bracing System of truss


A Pratt type roof truss was designed from the following data:

Design Data:

Span = 40 feet

Span-to-rise ratio (pitch) = 4:1

Rise = 10 feet

Bay distance (truss-to-truss distance) = 25 feet.

Exposure category: Exposure A

Truss is supported on brick wall of height = 12 feet.



Figure 9.4: Pratt type Roof Truss

Table: Member Specifications

Member type Design section Area (in2)

Top chord 4

13

2

13 L 1.56

Bottom chord 16

534 L 2.09

Verticals 16

3

2

12

2

12 L 0.9

Web member / Diagonals 16

3

2

123 L 0.996

Vertical bracing 4

1

2

11

2

11 L 0.688

Top chord bracing 4

1

4

31

4

31 L 0.813

Bottom chord strut 16

522 L 1.15

Purlin C 7×9.8 2.87

Sagrod #3 bar 0.11



Solution:

The various members of the truss network are designed as follows:

Table: Calculation of total weight of the truss members

Member

type

Shape of

Section Design section Quantity

Length

(ft)

Unit

Weight

(lb/ft)

Total

Weight

(lb)

Top chord Angle 4

13

2

13 L 12 7.453 5.40 40.246

Bottom

chord Angle

16

534 L 12 6.667 7.20 48.002

Verticals

Angle

16

3

2

12

2

12 L

4 3.333 3.10 10.332

Angle 4 6.667 3.10 20.668

Angle 2 10 3.10 31

Web

member /

Diagonals

Angle

16

3

2

123 L

4 7.454 3.39 25.269

Angle 4 9.429 3.39 31.964

Vertical

bracing Angle

4

1

2

11

2

11 L 2 26.926 2.34 63.007

Top chord

bracing Angle

4

1

4

31

4

31 L 4 29.107 2.77 80.626

Bottom

chord strut Angle

16

522 L 7 25 3.92 98

Purlin Channel C 7×9.8 5 25 9.8 245

Sagrod Cylindrical #3 bar 2 22.361 0.38 8.497

Total weight = 702.611 (lb)

Total weight of steel members = 702.611 lb

Unit Price = A tk/lb

Total Cost= 702.611 x A = X1 tk

Erection and Welding Cost = 10% of total cost= X2

Estimated cost=X1+X2 tk



Part 10: Uses of Software in Construction Estimation

10.1 Introduction

Measurement software applications, as a useful IT tool, can help a quantity surveyor to speed up the

measurement works. Studies have shown that workers (quantity surveyors) spent up to 80% of their

working time on measurement of quantities to produce bills of quantities (Keng & Ching, 2011).

The benefits of using measurement software in the preparation of Bill of Quantity (BoQ) are high

accuracy, easy to edit the BoQ, speed up measurement works, high traceability, user friendly, and

reduction of workforce to do measurement works. These indicate that the adoption and the use of

measurement software can bring improvement in editing BoQ, accuracy, traceability, and timeliness.

By using measurement software, it can enhance the service quality of quantity surveying firms and

enabled them to meet the high expectations of the clients.

10.2 Software Used in Quantity Surveying

A large number of software is available to the quantity surveyors. The two main functions of these

software include taking off measurement and preparing the BoQ. A few prominent software are

listed below:

1. QuickMeasure OS

2. PriMus-To

3. PlanSwift

4. On Center Takeoff

5. Bluebeam Revu

6. Vu360

7. Easy-Pro Builders Estimator

8. iScope

9. Estimate

10. PlanViewer

This manual discusses briefly on the usage of QuickMeasure OS.

10.3 QuickMeasure OS

QuickMeasure OS is one of the most popular software because of its user-friendliness, strong

support system and most importantly, its integration with a widely-used and available software

Microsoft Excel. A brief discussion on the use of this is provided below using the example, where

it is shown how to calculate volume of brick masonry.

Step One: Open Excel file. Create a new list identifying all the items of measurement and include

the additional columns required for complete calculations along with their formulae.



Figure 10.1: Listing Items

In this case, for each room, the columns created are: Length, Wall Height, Single/ double sided,

surface area, cost per unit surface area and total.

Step Two: The QuickMeasure Onscreen (QuickMeasure OS) loads automatically as an add-in in

excel. This add-in can be used in the form of OnScreen as well as Digitizer, as shown in Figure 10.2.

In this case we are using OnScreen version.

Step Three: Click on “Set Description Column Row”. This will open up a window (Figure 10.3)

where you need to enter the column or row number that includes your primary product description.

In this case, it is the Identity of the Rooms, so use column A.

List of all items of

measurements along

with cells reserved

for subsequent

calculations

QuickMeasure

Toolbar



Figure 10.2: Choosing OnScreen option

Figure 10.3: Setting the Description Column



Step four: Clicking on Manage OnScreen Images (Figure 10.4) will open up a window to load

images (Figure 10.5) in the form of jpg, gif, pdf, etc. As shown in Figure 10.6, click “Add” to load

more images, then click “Done” when you are finished adding images.

Figure 10.4: Clicking on Manage OnScreen Images

Figure 10.5: Adding Plans



Figure 10.6: Managing Plans

Step five: In this example we are going to find the length (linear footage) of wall by take-off and

then the surface area and cost of wall for each room mentioned in different rows. Select the Linear

Footage column (Column B), because this is the column that we will populate with values from take-

off. This is shown in figure 10.7.

Figure 10.7: Selecting Linear Footage Column



Selecting

Polyline option

from the

QuickMeasure

Toolbar

In the custom toolbar, you can work with linear distances using either polyline or segments. In this

example, we will select polyline, as shown by the mouse cursor in Figure 10.8.

Figure 10.8: Selecting Polyline option from the QuickMeasure Toolbar

Selecting the polyline, then pressing enter will open up the QuickMeasure OnScreen Window, as

shown in Figure 10.9.

Figure 10.9: QuickMeasure OnScreen Window



Step six: In figure 10.9, the Scale Line is deselected by default. We now select it to set the scale of

the diagram, as shown in Figure 10.10.

Figure 10.10: Check Marking the Scale Line

Figure 10.11: Scaling the Plan

The scale line is

check marked.

Scale Line

Room #103

Blue line used

to draw line

for scaling.



Selecting the “Scale Line” will prompt the user to draw a line on the building plan. Drag the mouse

cursor over the distance in the image file to show the distance to be scaled, as shown using the light

blue line in Figure 10.11. The line length and direction can be adjusted as necessary. When you are

satisfied, click on “OK Done” in the top toolbar shown in Figure 10.11. A window called “Set Scale”

then opens up, where the real distance is specified. In this case, the real distance is 32 feet 0 inches,

as shown in Figure 10.12. Press “OK” when done.

Figure 10.12: Setting the Scale

After selecting the correct scale, uncheck the tick mark of the scale line. If a new scale is required,

check the tick mark again to adjust scale.

Step seven: Select the first room (Room #103) out of the list and then click on “Line” on the top

toolbar in Figure 10.13. You can also change and predefine the line style using “Select Line Style”.

After you are satisfied, click on “Line” on the top toolbar. Use the mouse to define the length of the

walls of the room. In this case, blue line is used to define the walls of Room #103 as shown in Figure

10.14. Left click each time you want to set a point and double click when you are finished drawing

the line. The length of the line is automatically measured as show in Figure 10.15. You can navigate

around the image using Right-click button or the mouse scroll button. Once you are ready to move

to the next room, click on “OK Next”. You can change line style for the new room to differentiate

from the previous room. If you want to skip some rooms and jump ahead to, for example Room

#112, just select Room #112 from the list on the left side and digitize the respective room.



Figure 10.13: Selecting Room #103

Figure 10.14: Demarcating Walls for Take-off

Blue line used

to demarcate the

walls of Room

#103 and help in

take-off



Once you are finished with your digitizing, click “OK Done”. The Linear Footage (Column B) of

the excel file will be populated with the digitized measurements. Fill up the remaining columns

manually and finally get the total cost of materials for the walls as shown in figure 10.16.

Figure 10.15: Finishing the Take-off

Figure 10.16: Automatic Transfer of Measurements to Excel



References

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ce 208 quantity surveying

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